Proving a property by using the spectral theorem

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $(F,\langle\cdot,\cdot\rangle)$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$).

I want to show by using the spectral theorem that: for any $x\in F$ there exists a sequence $(x_n)_n$ with $$M^{1/2}x=\lim_{n\to \infty} Mx_n\;.$$

Answer 1

You can use the binomial expansion. Suppose that $0 \le M \le I$, which you can do by scaling $M$ by a positive constant, if necessary. Then the unique positive square root of $M$ is given by the binomial expansion, $$ M^{1/2} = (I-(I-M))^{1/2}=I-\sum_{n=1}^{\infty}c_n(I-M)^n, $$ where $c_n$ are positive constants such that $\sum_{n=1}^{\infty}c_n=1$. You can use that to show that the range of $M$ is dense in the range of $M^{1/2}$.

Using the spectral theorem, and assuming $L$ is an upper bound for the spectrum of $M$, $$ M\int_{1/n}^{L}\lambda^{-1/2}dE(\lambda)x= \int_{1/n}^{L}\lambda^{1/2}dE(\lambda)x. $$

So $x_n=\int_{1/n}^{L}\lambda^{-1/2}dE(\lambda)x$ is such that $Mx_n\rightarrow M^{1/2}x$ as $n\rightarrow\infty$, which proves that the range of $M$ is dense in the range of $M^{1/2}$.

Answer 2

There seems to be a misunderstanding of the claim on page 5. Given $M\geq0$ in $\mathcal{H}$, there is an associated Hilbert space $\mathbf{R}(M^{1/2})$, with the inner product \begin{align*} \left\langle x,y\right\rangle _{M} & :=\left\langle x,My\right\rangle \end{align*} modulo elements s.t. $\left\langle x,Mx\right\rangle =0$.

Claim: The range of $M$ is dense in $\mathbf{R}\left(M^{1/2}\right)$, i.e., \begin{align*} \left[\left\langle Mx,y\right\rangle _{M}=0,\;\forall x\in\mathcal{H}\right] & \Longrightarrow y=0\in\mathbf{R}(M^{1/2}). \end{align*} This is true, since $\left\langle Mx,y\right\rangle _{M}=0$, $\forall x\in\mathcal{H}$ iff $\left\langle x,M^{2}y\right\rangle =0$, $\forall x\in\mathcal{H}$ iff $M^{2}y=0$ in $\mathcal{H}$. Then \begin{align*} \left\langle y,My\right\rangle & \leq\left\Vert y\right\Vert ^{2}\left\langle y,M^{2}y\right\rangle =0, \end{align*} i.e., $y$ is the zero vector in $\mathbf{R}(M^{1/2})$.